Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
		import java.util.*;
		import java.io.*;
		class Main {
			static int n;
			static double arr[],dp[][];
			public static void main(String[] args) throws IOException{
				BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
				int t=1;
		//		Integer.parseInt(br.readLine().trim());
				StringBuilder printans=new StringBuilder();
				while(t-->0)
				{
					StringTokenizer tok=new StringTokenizer(br.readLine()," ");
					n=Integer.parseInt(tok.nextToken());
					tok=new StringTokenizer(br.readLine()," ");
					arr=new double[n];
					for(int i=0;i<n;i++) {
						arr[i]=Double.parseDouble(tok.nextToken());
					}
					dp=new double[n+1][n+1];
					memset();
					System.out.println(dfs(n,0));
					
					
				}
				
				
			}
			static double  dfs(int i,int h) {
				if(dp[i][h]!=-1d)return dp[i][h];
				if(i<=0) {
					if(h>n/2) {
						return 1;
					}
					return 0;
				}
				
				dp[i][h]=(double)(arr[i-1]*dfs(i-1,h+1)+(1-arr[i-1])*dfs(i-1,h));
				
				return dp[i][h];
				
				
				
			}
	//		
			static void memset() {
				for(double a[]:dp) {
					Arrays.fill(a,-1d);
				}
			}
			
				
		}